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Reptile [31]
3 years ago
11

The largest county in the state of Texas is 6064 square miles larger than the smallest county in the same state. The size of the

largest county is 48 times the size of the smallest county plus one square mile. How large is the smallest county in the state of Texas ?
Mathematics
1 answer:
Ann [662]3 years ago
7 0

The smallest county is 129 square miles large.

Step-by-step explanation:

Let,

Smallest county = x

Largest county = y

According to given statement;

The largest county in the state of Texas is 6064 square miles larger than the smallest county in the same state.

y = x+6064     Eqn 1

The size of the largest county is 48 times the size of the smallest county plus one square mile.

y = 48x+1         Eqn 2

As Eqn 1 and Eqn 2 are both equal to y, therefore,

Eqn 1 = Eqn 2

x+6064=48x+1\\6064-1=48x-x\\6063=47x\\47x=6063

Dividing both sides by 47

\frac{47x}{47}=\frac{6063}{47}\\x=129\ square \ miles

The smallest county is 129 square miles large.

Keywords: linear equation, substitution method

Learn more about substitution method at:

  • brainly.com/question/9328925
  • brainly.com/question/9184197

#LearnwithBrainly

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Jason runs 440 yards in 75 seconds.at this rate,how many minutes does it take him to run a mile​
Kazeer [188]

Answer:

5 minutes

Step-by-step explanation:

Do a proportion

440/75= 1760/x

cross multiply

440x=1760(75)

simplify

132000/440

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Nastasia [14]

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A student created this table to represent a linear relationship between x and y.X Y-2. 10.0-1. 7.50. 5.01. 2.52. 0The student sa
alex41 [277]

The relationship between x and y is represent as:

Since, the relationship is linear.

The standard form of equation of line is:

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

Consider any two set x and y values from the given relationship.

Let (-2, 10) and (-1,7.5)

\text{Substitute x}_1=-2,y_1=10,x_2=-1,y_2=7.5\text{ in the standard equation of line.}

\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1) \\ y-10=\frac{7.5-10}{-1-(-2)}(x-(-2)) \\ y-10=\frac{-2.5}{1}(x+2) \\ y-10=-2.5(x+2) \\ y=-2.5(x+2)+10 \end{gathered}

The equation of the linear relationship between x and y is:

y = -2.5(x + 2) + 10

Now, to check that the point (9, -17.5) lies on the represented relationship between x and y

Substitute x = 9 and y = -17.5 in the equation y = -2.5(x + 2) + 10

y = -2.5(x + 2) + 10

-17.5 = -2.5(9 + 2) + 10

-17.5 = -2.5(11) + 10

-17.5 = -27.5 + 10

-17.5 = -17.5

Thus, LHS = RHS

Hence the point (9, -17.5) lie on the given linear relationship between x and y.

Answer: The point (9, -17.5) lie on the given linear relationship between x and y.

8 0
1 year ago
Can anyone help with this using soh cah toa and rounding the answer to the nearest tenth of a foot?
Black_prince [1.1K]

Answer:

X=12.09

Step-by-step explanation:

Tangent=opposite/adjacent

Tan29=6.7/x

X tan29=6.7

X=6.7/tan29

X=12.087

X=12.09

7 0
3 years ago
According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var
-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

A)

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.7

10.9

11.2

11.5

11.8

n=10(even)

Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : \sqrt{\frac{\sum(x-\bar{x})^2}{n}}

Standard deviation :\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}

Standard deviation :0.89916

C)

8.3

,10.3

,10.5

,10.6

,10.7

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range :(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0
3 years ago
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